Math 202B: Lecture 14

When things get confusing, it helps to come back to basics. Let $V$ be a Hilbert space of dimension $1< n <\infty,$ and let $X \subset V$ be an orthonormal basis.

Definition 14.1. The matrix elements relative to $X$ are the linear functionals

$M_{xy} \colon \mathcal{L}(V) \longrightarrow \mathbb{C}, \quad (x,y) \in X \times X$

defined by

$M_{xy}(A) =\langle x, Ay \rangle, \quad A \in \mathcal{L}(V).$

Proposition 14.1. Two operators $A,B \in \mathcal{L}(V)$ are equal if and only if $M_{xy}(A) = M_{xy}(B)$ for all $x,y \in X.$

Problem 14.1. Prove Proposition 14.1.

Definition 14.2. The trace on $\mathcal{L}(V)$ relative to $X$ is the linear functional defined by

$\mathrm{Tr}_X = \sum\limits_{x \in X} M_{xx}.$

Thus,

$\mathrm{Tr}_X(A) = \sum\limits_{x \in X} \langle x,Ax\rangle, \quad A \in \mathcal{L}(V).$

Proposition 14.2. We have

$\mathrm{Tr}_X(I) = \dim V;$
$\mathrm{Tr}_X(AB)=\mathrm{Tr}_X(BA)$ ;
$\mathrm{Tr}_X(A^*A) \geq 0$ with equality if and only if $A=0.$

Proof: First, we have

$\mathrm{Tr}_X(I) = \sum\limits_{x \in X} \langle x,Ix\rangle = \sum\limits_{x\in X}1 = |X|=\dim V.$

Second, we have

$\mathrm{Tr}_X(AB)=\sum\limits_{x\in X} \langle x,ABx\rangle=\sum\limits_{x \in X} \langle x,A\sum\limits_{y \in X} \langle y,Bx\rangle y\rangle=\sum\limits_{x \in X}\sum\limits_{y\in X} \langle x,Ay\rangle \langle y,Bx \rangle,$

and changing order of summation this is

$\sum\limits_{y\in X}\sum\limits_{x \in X}\langle y,Bx \rangle\langle x,Ay\rangle=\sum\limits_{y \in X} \langle y,B\sum\limits_{x \in X} \langle x,Ay\rangle x\rangle=\sum\limits_{y \in X} \langle y,BAy\rangle=\mathrm{Tr}_X(BA).$

Third, we have

$\mathrm{Tr}_X(A^*A) = \sum\limits_{x \in X} \langle x,A^*Ax\rangle = \sum\limits_{x \in X} \langle Ax,Ax\rangle = \sum\limits_{x\in X} \|Ax\|^2,$

which vanishes if and only if every term of the sum vanishes, and this occurs if and only if $A$ maps every vector in the basis $X$ to the zero vector.

-QED

Note that the definition of $\mathrm{Tr}_X$ makes sense even if the basis $X$ is not orthonormal. We claim that Proposition 14.2 remains valid in this case. Proof: define a new scalar product on $V$ in which $X$ is orthonormal.

Proposition 14.3. Let $X$ and $Y$ be two orthonormal bases of $V$ . We have $\mathrm{Tr}_X=\mathrm{Tr}_Y$ .

Proof: For any enumeration $x_1,\dots,x_n$ of $X$ , and any enumeration $y_1,\dots,y_n$ of $Y$ , the linear operator $U \colon V \to V$ defined by $Ux_i=y_i$ is unitary. We have

$\mathrm{Tr}_Y(A) = \sum\limits_{i=1}^n \langle y_i,Ay_i\rangle=\sum\limits_{I=1}^n \langle Ux_i,AUx_i\rangle = \mathrm{Tr}_X(U^*AU).$

By Proposition 14.2, we have $\mathrm{Tr}_X(U^*AU)=\mathrm{Tr}_X(AUU^*)=\mathrm{Tr}_X(A).$

-QED

Proposition 14.3 again holds without assuming orthonormality of $X,Y,$ for the same reason as above. We have thus shown that Definition 14.3 gives the same linear functional on $\mathcal{L}(V)$ no matter what basis $X$ is used to define it, and we call this functional the trace on $V.$

Definition 14.3. The spectrum of $A \in \mathcal{L}(V)$ is the subset of $\mathbb{C}$ defined by

$\sigma(A) = \{\alpha \in \mathbb{C} \colon \alpha I - A \text{ not invertible}\}.$

The numbers in $\sigma(A)$ are called the eigenvalues of $A.$

Theorem 14.4. (Fundamental Theorem of Algebra) Every $A \in \mathcal{L}(V)$ has nonempty spectrum $\sigma(A).$ That is, every operator has an eigenvalue.

Problem 14.2. Prove that, for any $A \in \mathcal{L}(V)$ , we have

$\mathrm{Tr} A = \sum\limits_{\alpha \in A} \alpha.$

Note Definition 14.3 makes sense in any algebra $\mathcal{A}$ , though the term “eigenvalue” is only used in the case $\mathcal{A}=\mathcal{L}(V).$

Problem 14.3. What is the spectrum of a function $A \in \mathcal{F}(X)$ ?

One of the most important aspects of the trace on $\mathcal{L}(V)$ is that it gives us a scalar product on $\mathcal{L}(V)$ which interfaces well with its algebra structure.

Definition 14.4. The Frobenius scalar product on $\mathcal{L}(V)$ is defined by $\langle A,B \rangle = \mathrm{Tr} A^*B.$

The definition of the Frobenius scalar product should be familiar from Math 202A. The associated norm is called the Frobenius norm, $\|A\|=\sqrt{\mathrm{Tr A^*A}}.$

Problem 14.4. Show that the Frobenius scalar product is indeed a scalar product, and moreover that it interacts with multiplication and conjugation in $\mathcal{L}(V)$ according to $\langle AB,C\rangle =\langle B,A^*C\rangle,$ for all $A,B,C \in \mathcal{L}(V).$

Definition 14.5. The elementary operators relative to $X$ are the operators $E_{xy} \in \mathcal{L}(V)$ defined by

$E_{xy}z = x \langle y,z\rangle, \quad x,y,z \in X.$

From the definition, we have

$M_{wz}(E_{xy})=\langle w,E_{xy}z\rangle =\langle w,x\langle y,z\rangle\rangle=\langle w,x \rangle \langle y,z \rangle, \quad x,y,z,w \in X.$

Problem 14.4. Show that the trace of an elementary operator is $\mathrm{Tr} E_{xy}=\langle x,y \rangle.$

Proposition 14.4. The elementary operators satisfy

$E_{xy}^*=E_{yx} \quad\text{and}\quad E_{wx}E_{yz}=\langle x,y\rangle E_{wz}, \quad w,x,y,z \in X.$

Proof: For any $w,x,y,z \in X,$ we have

$\langle w,E_{xy}z\rangle = \langle w,x\langle y,z\rangle\rangle=\langle w,x\rangle \langle y,z\rangle$

and

$\langle E_{yx}w,z\rangle = \langle y\langle x,w\rangle,z\rangle = \langle x,w\rangle \langle y,z \rangle,$

so indeed $E_{xy}^*=E_{yx}.$

Now consider the $v,a \in X$ and consider the matrix element

$M_{va}(E_{wx}E_{yz})=\langle v,E_{wx}E_{yz}a\rangle =$

-QED

Theorem 14.5. The elementary operators $\{E_{xy}\colon (x,y) \in X \times X\}$ form an orthonormal basis of $\mathcal{L}(V)$ relative to the Frobenius scalar product.

Proof: First we check that $\{E_{xy}\colon (x,y) \in X \times X\}$ is an orthonormal set with respect to the Frobenius scalar product. For any $w,x,y,z \in X$ we have

$\langle E_{wx},E_{yz}\rangle = \mathrm{Tr}(E_{wx}^*E_{yz})=\mathrm{Tr}(E_{xw}E_{yz}) = \langle w,y\rangle\mathrm{Tr}(E_{xz})=\langle w,y\rangle \langle x,z\rangle.$

Now we check that $\{E_{xy}\colon (x,y) \in X \times X\}$ spans $\mathcal{L}(V).$ Let $A \in \mathcal{L}(V)$ be an arbitrary operator, and write $\alpha_{xy} = \langle x,Ay\rangle$ for its matrix elements relative $X.$ We claim that

$A = \sum\limits_{(x,y) \in X\times X} \alpha_{xy}E_{xy}.$

Let $R$ be the operator on the right hand side of the above equation. Then, the matrix elements of $R$ are

$\langle w,Rz\rangle = \sum\limits_{(x,y) \in X \times X} \alpha_{xy}\langle w,E_{xy}z\rangle \sum\limits_{(x,y) \in X\times X} \alpha_{xy}\langle w,x\rangle \langle y,z\rangle=\alpha_{wz},$

which shows that all matrix elements of $A$ and $R$ are equal, so that these operators are the same by Proposition 14.1.

-QED

Unlike the elementary basis $\{E_x \colon x \in X\}$ of the function algebra $\mathcal{F}(X)$ , the elementary basis $\{E_{xy} \colon x,y \in X\}$ of the linear algebra $\mathcal{L}(V)$ is not a basis of orthogonal projections — indeed, $\mathcal{L}(V)$ is not a commutative algebra, so it does not admit a basis of orthogonal projections. However, the diagonal elementary operators $\{E_{xx} \colon x\in X\}$ are orthogonal projections spanning a commutative subalgebra of $\mathcal{L}(V)$ isomorphic to $\mathcal{F}(X)$ . Indeed, the isomorphism is simply

$E_{xx} \mapsto E_x, \quad x \in X.$

The trace on $\mathcal{L}(V)$ is unique in the following sense.

Theorem 14.5. If $\tau \colon \mathcal{L}(V) \to \mathbb{C}$ is a linear functional which satisfies $\tau(AB)=\tau(BA)$ for all $A,B \in \mathcal{L}(V)$ , then $\tau$ is a scalar multiple of the trace.

Proof: Since $\tau$ is linear, it is uniquely determined by its values on the elementary basis $E_{xy}$ of $\mathcal{L}(V)$ .

For distinct $x,y \in X$ , we have

$\tau(E_{xy})=\tau(E_{xy}E_{yy}) = \tau(E_{yy}E_{xy})=\tau(0)=0.$

For the diagonal elementary operators,

$\tau(E_{xx})=\tau(E_{xy}E_{yx})=\tau(E_{yx}E_{xy})=\tau(E_{yy}).$

Thus, $\tau$ is constant on the diagonal elementary operators, i.e. there is a constant $\gamma$ such that $\tau(E_{xx})=\gamma$ for all $x \in X.$ Thus for any